ML Aggarwal Class 12 Maths Solutions Section A Chapter 10 Probability Ex 10.9

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ML Aggarwal Class 12 Maths Solutions Section A Chapter 10 Probability Ex 10.9

Question 1.
Is it possible that the mean of a binomial distribution is 15 and its standard deviation is 5 ?
Answer:
Here mean of binomial distribution = 15 = np …(1)
and standard deviation of B.D = σ = 5 = \(\sqrt{n p q}\)
∴ npq = 25 …………(2)
From (1) and (2); we have
15 × 9 = 25 ⇒ q = \(\frac{25}{15}=\frac{5}{3}\) > 1
Since prob. of an event can’t be greater than 1.
Such a binomial distribution is not possible.

Question 2.
Find the mean of the Binomial distribution B(8, \(\frac{1}{4}\))?
Answer:
Given binomial distribution be B(8, \(\frac{1}{4}\))
Here n = 8 ; P = \(\frac{1}{4}\)
∴ q = 1 – p = 1 – \(\frac{1}{4}\) = \(\frac{3}{4}\)
∴ Mean of binomial distribution
= np = 8 × \(\frac{1}{4}\) = 2

Question 3.
In a binomial distribution, if n = 20 and q = 0.6, then find its mean.
Answer:
Given n = 20 ; q = 0.6
p = 1 – q = 1 – 0.6 = 0.4
Mean = np = 20 × 0.4 = 8

Question 4.
In a binomial distribution, if mean is 5 and variance is 4, then find the number of trials.
Answer:
Here mean = np = 5 …(1)
variance = npq = 4 …(2)
from (1) and (2) ; we have
q = \(\frac{4}{5}\) ∴ p = 1 – q = 1 – \(\frac{4}{5}\) = \(\frac{1}{5}\)
from(1) ; n × \(\frac{1}{5}\) = 5 ⇒ n = 25

Question 5.
If the mean of a binomial distribution is 20 and its standard deviation is 4, then find the value of p.
Answer:
Given mean = np = 20 …………(1)
and σ = S.D = \(\sqrt{n p q}\) = 4
⇒ npq = 16
from (1) and (2); we have
20 × q = 16
⇒ q = \(\frac{4}{5}\)
∴ p = 1 – q = 1 – \(\frac{4}{5}=\frac{1}{5}\)

Question 6.
Find the expectation of number of heads in 20 tosses of a fair coin.
Answer:
Let p = prob. of getting head in single toss of coin = \(\frac{1}{2}\)
q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\) and n = 20
Then expetation or mean of binomial distribution = np
= 20 × \(\frac{1}{2}\) = 10

Question 7.
A binomial distribution has n = 60 and p = 0-7. Compute the (/) mean and ‘ (//) standard deviation.
Answer:
Given n = 60; p = 0.7
q = 1 – p = 1 – 0.7 = 0.3
∴ Mean of Binomial distribution = np = 60 × 0.7 = 42
and standard derivation of B.D = σ = \(\sqrt{n p q}\)
= \(\sqrt{60 \times 0.7 \times 0.3}=\sqrt{12.6}\)
= 3.55

Question 8.
If the probability of a defective bolt is 0-1, then find the mean and the standard deviation for the number of defective bolts in a total of 400 bolts.
Answer:
Let p = probability of defective bolt = 01
Here n = 400 and q = 1 – p = 1 – 0.1 = 0.9
Then mean of binomial distribution = np = 400 × 0.1 = 40
and Variance of B.D
= npq = 400 × 0.1 × 0.9 = 36
∴ S.D. of binomial distribution = \(\sqrt{36}\) = 6

Question 9.
In a binomial distribution, if mean and variance are 4 and 2 respectively, then find the probability of exactly one success.
Answer:
Given mean = np = 4 …(1)
variance = npq = 2 …(2)
from (1) and (2); we have
4q = 2 ⇒ q = \(\frac{1}{2}\)
from(1); n x \(\frac{1}{2}\) = 4 ⇒ n = 8

Thus by binomial distribution, we have
P (X = r) = ⁿC_rp^r q^n-r
= ⁸C_r\(\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{8-r}\)
= ⁸C_r\(\left(\frac{1}{2}\right)^8\)

Prob. of exactly one success = P (X = 1)
= ⁸C₁\(\left(\frac{1}{2}\right)^8=\frac{8}{2^8}=\frac{1}{32}\)

Question 10.
A die is tossed twice. A success is defined as getting an odd number on a random toss. Find the mean and variance of the number of successes.
Answer:
In single throw of die, total no. of outcomes = 6
p = prob. of getting an odd number = \(\frac{3}{6}=\frac{1}{2}\)
∴ q = 1 – p = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\); n = 2
Then by binomial distribution, we have
Mean = np = 2 × \(\frac{1}{2}\) = 1
Variance = npq
= 2 × \(\frac{1}{2} \times \frac{1}{2}=\frac{1}{2}\)

Question 11.
If X follows binomial distribution with parameters n = 5, p and P (X = 2) = 9 P (X = 3), then find the value of p.
Answer:
Given P (X = 2) = 9P (X = 3) …(1)
We know that,
P (X = r) =ⁿC_r p^r q^{n – r} = ⁵C_r p^r q^{5 – r}
from (1) ; ⁵C₂ p² q³ = 9 . ⁵C₃ p³ q²
⇒ p²q³ = 9p³q² ⇒ q = 9p
Since p + q = 1 ⇒ 10p = 1
⇒ p = \(\frac{1}{10}\) q = \(\frac{9}{10}\)

Question 12.
Three cards are drawn successively with replacement from a well-shuffled pack of 52 cards. Find the mean and the variance of the number of red cards drawn.
Answer:
Let p = prob. of drawing a red card
= \(\frac{26}{52}=\frac{1}{2}\)
q = 1 – p = 1 – \(\frac{1}{2}=\frac{1}{2}\)
Since cards are drawn successively with replacement so it is a problem of binomial distribution. Here rt = 3
∴ Mean = np =3 × \(\frac{1}{2}=\frac{3}{2}\)
and variance = npq = 3 × \(\frac{1}{2} \times \frac{1}{2}=\frac{3}{4}\)

Question 13.
A fair die is thrown two times. Find the probability distribution of the number of sixes. Hence, find the mean of the number of sixes.
Answer:
Since a dice is thrown no. of times so events are independent. Hence it is a problem of binomial distribution.
Let p = prob. of getting 6 = \(\frac{1}{6}\)
q = 1 – p = 1 – \(\frac{1}{6}=\frac{5}{6}\)
Here n = 2
∴ Mean = np = 2 × \(\frac{1}{6}=\frac{1}{3}\)
Let X denotes no. of sixes drawn so X can take values 0, 1, 2
We know that P(X = r) = ⁿC_r p^r q^{n – r}
= ²C_r\(\left(\frac{1}{6}\right)^r\left(\frac{5}{6}\right)^{2-r}\)

Thus probability distribution of X be given by

Question 13(Old).
Five bad oranges are accidently mixed with 20 good ones. If four oranges are drwan one v \ by one successively with replacement, then find the probability distribution of number of bad oranges drawn. Hence, find the mean and variance of the distribution.
Answer:
Let p = probability of getting bad orange = \(\frac{5}{25}=\frac{1}{5}\)
Then q = 1 – p = 1 – \(\frac{1}{5}=\frac{4}{5}\) and n = 4
Let X denotes the number of bad oranges drawn in 4 draws with replacement.

Then by binomial distribution, we have

the probability distribution of X is given below:

Aliter : Mean of binomial distribution -np = 4 × \(\frac{1}{5}=\frac{4}{5}\)

and Variance of binomial distribution = npq = 4 × \(\frac{1}{5} \times \frac{4}{5}=\frac{16}{25}\)

Question 14.
A die is thrown 20 times. Getting a number greater than 4 is considered a success. Find the mean and the variance of the number of successes.
Answer:
Here n = 20, p = prob. of getting a number >4 = \(\frac{2}{6}=\frac{1}{3}\)
q = 1 – p = 1 –\(\frac{1}{3}=\frac{2}{3}\)
Mean of B.D – np = 20 × \(\frac{1}{3}=\frac{20}{3}\)
and variance of B.D = npq = 20 × \(\frac{1}{3} \times \frac{2}{3}=\frac{40}{9}\)

Question 15.
A die is thrown 3 times. Let X be ‘the number of two’s seen’, then find the \ V probability distribution of X. Also find the expectation of X. (NCERT Exemplar)
Answer:
When a die is tossed, the possible outcomes are {1,2, 3, 4, 5, 6} and all outcomes are equally likely.
So it is case of binomial distribution.
p = prob. of a number > 4 = \(\frac{2}{6}=\frac{1}{3}\)
∴ q =1 – p = 1 – \(\frac{1}{3}=\frac{2}{3}\) > here n = 2

Then by binomial distribution, we have

The probability distribution of X is given below:

∴ Mean of binomial distribution
= np = 2 × \(\frac{1}{3}=\frac{2}{3}\)
and Variance of B.D = npq = 2 × \(\frac{1}{3} \times \frac{2}{3}=\frac{4}{9}\)
and S.D of B.D = \(\sqrt{\frac{4}{9}}=\frac{2}{3}\)

Question 16.
Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as ‘a number greater than 4’. Also find the mean, variance and standard deviation of the distribution.
Answer:
Here n = 2,
p = probability of getting a number > 4
= \(\frac{2}{6}=\frac{1}{3}\)
q = 1 – p = 1 – \(\frac{1}{3}=\frac{2}{3}\)
Thus mean = np = 2 × \(\frac{1}{3}=\frac{2}{3}\)

and variance = npq = σ² = 2 × \(\frac{1}{3} \times \frac{2}{3}=\frac{4}{9}\)
and S.D. = σ = \(\sqrt{\frac{4}{9}}=\frac{2}{3}\)

Question 17.
Two cards are drawn successively with replacement from a well-shuffled pack of 52 cards. Find the probability distribution of the number of aces drawn. Also find the mean and standard deviation of the distribution.
Answer:
Since cards are drawn successively with replacement so events are independent.
Hence it is a case of binomial distribution.
p = prob. of getting an ace = \(\frac{4}{52}=\frac{1}{13}\)
q = 1 – p = 1 – \(\frac{1}{13}=\frac{12}{13}\)
Here n = 2.
Let X be the binomial variate and denote the no. of aces in 2 draws.
∴ X can takes values 0, 1,2.

Thus, the probability distribution of X is given below :

Mean of binomial distribution = np = 2 × \(\frac{1}{13}=\frac{2}{13}\)
Variance of B.D = σ² = npq – 2 × \(\frac{1}{13} \times \frac{12}{13}=\frac{24}{169}\)
S.D of B.D = σ = \(\sqrt{\frac{24}{169}}=\frac{2 \sqrt{6}}{13}\)

Question 17(Old).
The mean and the variance of a binomial distribution are 4 and 2 respectively. Find the probability of:
(i) exactly one success
(ii) exactly two successes.
Answer:
Since mean of binomial distribution = 4 = np …(1)
and Variance of binomial distribution = 2 = npq …(2)
from (1) and (2) ; we have
2 = 4q ⇒ q = \(\frac{1}{2}\)
∴ p = 1 – q = 1 – \(\frac{1}{2}\) = \(\frac{1}{2}\)
from (1); \(\frac{n}{2}\) = 4 ⇒ n = 8
Then by binomial distribution we have
P(X = r) = ⁿC_r p^r q^{n – r}
= ⁸C_r\(\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{8-r}\)
⇒ P(X = r) = ⁸C_r\(\left(\frac{1}{2}\right)^8\)

(i) required probability of getting exactly one success
= P(X= 1) = ⁸C₁\(\left(\frac{1}{2}\right)^8=\frac{8}{256}=\frac{1}{32}\)

(ii) required probability of getting exactly two successes
= P(X = 2) = ⁸C₂\(\left(\frac{1}{2}\right)^8=\frac{8 \times 7}{2} \times\left(\frac{1}{2}\right)^8\)
= \(\frac{28}{256}=\frac{7}{64}\)

Question 18.
A die is tossed three times. Getting an even number is considered a success. Find the binomial distribution. Also find the variance of the distribution.
Answer:
p = probability of success i.e. getting an even number in a single toss of die
= \(\frac{3}{6}=\frac{1}{2}\) {2, 4, 6}
q = 1 -p = 1 – \(\frac{1}{2}=\frac{1}{2}\)
Here, the given distribution be binomial distribution with n = 3 since all trials are independent of each other.
∴ required binomial distribution be given by (q + p)ⁿ i.e. \(\left(\frac{1}{2}+\frac{1}{2}\right)^3\)
∴ Variance of B.D = npq = 3 × \(\frac{1}{2} \times \frac{1}{2}=\frac{3}{4}\)

Question 19.
Find the mean, variance and sta.dard deviation of the number of doublets is three throws of a pair of dice.
Answer:
p = prob. of getting a doublet in throwing of pair of dice = \(\frac{6}{36}=\frac{1}{6}\)
[Here favourable cases are ; (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)]
q = 1 – p = 1 – \(\frac{1}{6}=\frac{5}{6}\)
Let X denotes the no. of doublets so X can take values 0, 1, 2, 3 with n 3
Then by binomial distribution, mean np =3 × \(\frac{1}{6}=\frac{1}{2}\)
and variance a npq = 3 × \(\frac{1}{6} \times \frac{5}{6}=\frac{5}{12}\)
and S.D = σ = \(\sqrt{n p q}=\sqrt{\frac{5}{12}}\)

Question 20.
An urn contains 3 white and 6 red bails. Four balls are drawn one by one with replacement from the urn. Find the probability distribution of the number of red balls drawn. Also find the mean and variance of the distribution.
Answer:
Here total numberof balls = 3 + 6 = 9 and no. of red balls = 6
Let p = probability of drawing a red ball = \(\frac{6}{9}=\frac{2}{3}\)
q = 1 – p = 1 – \(\frac{2}{3}=\frac{1}{3}\) and n = 4
Let X denotes the number of red balls in 4 draws with replacement.
Then by binomial distribution, we have

The probability distribution of X is given below :

∴ Mean = ΣXP(X)
= 0 × \(\frac{1}{81}\) + 1 × \(\frac{8}{81}\) + 2 × \(\frac{24}{81}\) + 3 × \(\frac{32}{81}\) + 4 × \(\frac{16}{81}=\frac{216}{81}=\frac{8}{3}\)
or Mean of binomial distribution = np = 4 × \(\frac{2}{3}=\frac{8}{3}\)

Question 21.
If the mean and variance of a binomial distribution are respectively 9 and 6, then find the distribution. r)\
Answer:
Let X be the binomial variate with parameters n and p.
Then mean = 9 and variance = 6
np = 9 ……..(1) and npq =6 …(2)

where r = 0, 1, 2, …………..27 be the required binomial distribution.

Question 21(Old).
If the sum and the product of the mean and variance of a binomial distribution ° are 24 and 128 respectively, then find the distribution.
Answer:
Since mean of B.D = np ;
Variance of B.D = npq
given sum of mean and variance of B.D be 24.
∴ np + npq = 24
⇒ np (1 + q) – 24 …(1)
also, product of mean and variance ofB.D = 128
(nP) (npq) =128
⇒ n²p²q² = 128 ……..(2)
On squaring (1) ; we have
n²p²(1 + q)² = (24)² = 576

On dividing (3) by (2); we have
⇒ \(\frac{(1+q)^2}{q}=\frac{576}{128}=\frac{9}{2}\)
⇒ 2(1 + q)² – 9q = 0
⇒ 2q² + 4q + 2 – 9q = 0
⇒ 2q² – 5q + 2 = 0
⇒ (q – 2) (2q – 1) = 0
⇒ q = 2, \(\frac{1}{2}\)
since 0 ≤ q ≤ 1 ∴ q = \(\frac{1}{2}\)
from (1); n x \(\frac{1}{2}\)(1 + \(\frac{1}{2}\)) = 24
⇒ \(\frac{n}{2} \times \frac{3}{2}\) = 24
⇒ n = \(\frac{24 \times 4}{3}\) = 32
Thus required binomial distribution is given by (q + p)ⁿ i.e. \(\left(\frac{1}{2}+\frac{1}{2}\right)^{32}\)

Question 22.
Find the binomial distribution for which the mean is 4 and variance 3.
Answer:
According to given, we have
np = mean = 4 …(1)
and npq = Variance = 3 ………(2)
From (1) and (2); we have
n × 4 = 3 ⇒ q = \(\frac{3}{4}\)
p = 1 – q = 1 – \(\frac{3}{4}=\frac{1}{4}\)
from (1); n × \(\frac{1}{4}\) = 4 ⇒ n – 16
Thus, required binomial distribution is given by
(q + p)ⁿ i.e. \(\left(\frac{3}{4}+\frac{1}{4}\right)^{16}\)

Question 23.
The mean number of successes of a binomial distribution is 240 and the cpp standard deviation is 12. Calculate the parameters of the distribution.
Answer:
Given, mean of B.D. = np = 240 …(1)
and S.D of B.D. = σ = \(\sqrt{n p q}\) =12
⇒ npq = (12)² = 144 …(2)

From (1) and (2) ; we have
240 × q = 144 ⇒ q = \(\frac{144}{240}=\frac{6}{10}=\frac{3}{5}\)
p = 1 – q = 1 – \(\frac{3}{5}=\frac{2}{5}\)

Thus from (1); n × \(\frac{2}{5}\) = 240
⇒ n = \(\frac{240 \times 5}{2}\) = 600
Thus the parameters of distribution are
n = 600
p = \(\frac{2}{5}\) and q = \(\frac{3}{5}\)

Question 24.
If X follows a binomial distribution with mean 4 and variance 2, then find P (X > 5).
Answer:
Given mean = np = A …(1)
and variance = npq = 2 …(2)
from (1) and (2); we have
from(1); n.\(\frac{1}{2}\) = 4 => n = 8

Thus by binomial distribution,
P(X = r) = ⁿC_r p^r q^{n – r}
= ⁸C_r\(\left(\frac{1}{2}\right)^r\left(\frac{1}{2}\right)^{8-r}\)
= ⁸C_r\(\left(\frac{1}{2}\right)^8\)

P (X ≥ 5) = P (X = 5) + P (X = 6) + P (X = 7) + P (X = 8)

Question 25.
If X follows a binomial distribution with mean 3 and variance \(\frac{3}{2}\) then find P(X ≤ 5).
Answer:
Let n and p are the parameters of binomial distribution.
Then mean = np and variance = npq
np =3 …(1)
and npq = \(\frac{3}{2}\) …(2)

On dividing eqn. (2) by eqn. (1); we have 3

Then binomial distribution is given by

Question 26.
If the sum of the mean and the variance . of a binomial distribution for 54 trials is 30, then find the distribution.
Answer:
Given mean of B.D. = np
and Variance of B.D = npq
according to given, we have
np + npq = 30 …(1)
also given n = 54 ∴ from (1); we have

Thus required B.D is given by (q + p)ⁿ
i.e \(\left(\frac{2}{3}+\frac{1}{3}\right)^{54}\)

Question 27.
If the sum of the mean and the variance of a binomial distribution for 5 trials is 1.8, then find the distribution.
Answer:
Given sum of mean and variance of binomial distribution for 5 trials is 1-8.
Mean of B.D = np ; Variance of B.D = npq
np + npq = 1.8
⇒ n (p + pq) = 1.8 …(1)
also n = 5 from (1) ; we have
5p (1 + q) = 18